The energy needed to keep a 75-watt lightbulb burning for 1.0 h is 270 kj. calculate the energy required to keep the lightbulb burning for 5.0 h in each of the following energy units. a) joules b) kilocalories

(a) 1.35 x 10^6 Joule

(b) 321.45 Kcal

Explanation:

Energy consumed in 1 hour = 270 kJ

So, Energy consumed in 5 hour = 270 x 5 = 1350 kJ

(a) Energy in joules = 1350 x 1000 J = 1.35 x 10^6 J

(b) 4.2 Joule = 1 calorie

So, 1.35 x 10^6 Joule = 1.35 x 10^6 / 4.2 = 0.32 x 10^6 Calorie

                                                                =  0.32 x 1000 Kcal = 321.45 Kcal

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j